Define $f : \mathbb{R}^2 \to \mathbb{R}$ by $f(0, 0) = 0$ and $f(x, y) = \frac{x^2y}{x^4+y^2}$ if $(x, y) \neq (0, 0)$. Show that all directional derivatives exist.
On page 44 of Analysis on Manifolds by Munkres he claims that for $u = (h, k) \neq (0, 0)$ we have $f_{(0, 0)}'(u) = \frac{h^2}{k}$ if $k \neq 0$ and $f_{(0, 0)}'(u) = 0$ if $k = 0$.
But in trying to arrive at this result on my own I ended up with the following result which I'll show now below. Let $u = (h, k) \neq (0, 0)$ then $$\frac{f((0, 0) + tu) -f(0, 0)}{t} = \frac{h^2k}{t^2h^4 + k^2}$$ and we have $$\lim_{t \to 0}\ \frac{h^2k}{t^2h^4 + k^2} = \frac{h^2}{k}$$
Now if I define $\phi(h, k) = \frac{h^2}{k}$ we can see that the directional derivative of $f$ at $0$ in the direction $(h, k)$ is $f_{(0, 0)}'(h, k) = \frac{h^2}{k}$, but then we can see clearly that $f_{(0, 0)}'$ is not defined for any $\alpha \in \mathbb{R}^2$ such that $\alpha = (h, 0)$, since the domain of $f_{(0, 0)}'$ is $\mathbb{R}^2$.
I don't see how this doesn't contradict Munkres' result. What have I done wrong? Furthemore how do I arrive at the result that he asserts above?
You are right, but $f_{(0, 0)}'(h, k) = \frac{h^2}{k}$ is true only for those $(h,k)$ such that $k \ne 0$.
For vectors of the form $(h,0)$ we have:
$$f'_{(0,0)}(h,0) = \lim_{t\to 0} \frac{f(t(h,0)) - f(0,0)}{t} = 0$$
as you already stated.
Therefore $f'_{(0,0)} : \mathbb{R}^2 \setminus \{0\}$ is defined as
$$f'_{0,0}(h,k) = \begin{cases} \frac{h^2}k, & \text{if $k \ne 0$} \\ 0, & \text{if $k = 0$} \end{cases}$$